Euler Identity Algebra Problem

**VinceW** · August 29th 2011, 07:28 PM

Using Euler's identity, find real positive constants $c$ and $\phi$ for all real $t$ such that:

$3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)$

I tried substituting the Euler formula for all sine and cosine values to get:

$\frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})$

Which suggests:

$ce^{j\phi} = 3 - 4 e^{j\pi /4}$

Which I don't see a solution to. Any ideas?

**Prove It** · August 29th 2011, 09:13 PM

Originally Posted by VinceW

Using Euler's identity, find real positive constants $c$ and $\phi$ for all real $t$ such that:

$3 \cos (2t) - 4 \sin (2t + \frac{\pi}{4}) = c \cos (2t + \phi)$

I tried substituting the Euler formula for all sine and cosine values to get:

$\frac{3}{2}(e^{j2t}+e^{-j2t}) - \frac{4}{2}(e^{j2t}e^{j\pi /4}-e^{-j2t}e^{-j\pi /4}) = \frac{c}{2}(e^{j2t}e^{j\phi}+e^{-j2t}e^{-j\phi})$

Which suggests:

$ce^{j\phi} = 3 - 4 e^{j\pi /4}$

Which I don't see a solution to. Any ideas?

$\displaystyle \begin{align*} 3\cos{(2t)} - 4\sin{\left(2t + \frac{\pi}{4}\right)} &= 3\cos{(2t)} - 4\left[\sin{(2t)}\cos{\left(\frac{\pi}{4}\right)} + \cos{(2t)}\sin{\left(\frac{\pi}{4}\right)}\right] \\ &= 3\cos{(2t)} - 4\left[\frac{\sqrt{2}}{2}\sin{(2t)} + \frac{\sqrt{2}}{2}\cos{(2t)}\right] \\ &= 3\cos{(2t)} - 2\sqrt{2}\sin{(2t)} - 2\sqrt{2}\cos{(2t)} \\ &= \left(3 - 2\sqrt{2}\right)\cos{(2t)} - 2\sqrt{2}\sin{(2t)} \end{align*}$

And...

$\displaystyle \begin{align*} c\cos{\left(2t + \phi\right)} &= c\left[\cos{(2t)}\cos{(\phi)} - \sin{(2t)}\sin{(\phi)}\right] \\ &= c\cos{(\phi)}\cos{(2t)} - c\sin{(\phi)}\sin{(2t)} \end{align*}$

So

$\displaystyle c\cos{(\phi)} = 3 - 2\sqrt{2} \\ \implies c^2\cos^2{(\phi)} = \left(3 - 2\sqrt{2}\right)^2 = 9 - 12\sqrt{2} + 8 = 17 - 12\sqrt{2}$

and $\displaystyle c\sin{(\phi)} = 2\sqrt{2} \implies c^2\sin^2{(\phi)} = 8$ .

Therefore

$\displaystyle \begin{align*} c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2} \\ c^2\left[\cos^2{(\phi)} + \sin^2{(\phi)} \right] &= 17 - 10\sqrt{2} \\ c^2(1) &= 17 - 10\sqrt{2} \\ c^2 &= 17 - 10\sqrt{2} \\ c &= \pm \sqrt{17 - 10\sqrt{2}} \end{align*}$

Can you go from here to get values for $\displaystyle \phi$ ?

**VinceW** · August 30th 2011, 09:26 AM

Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

I believe you made a small algebra error in that this:

$c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}$

should be:

$c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8$

thanks

**Prove It** · August 30th 2011, 10:04 AM

Originally Posted by VinceW

Thanks Prove It. You definitely solved it. You didn't use the Euler formula, but I guess that wasn't applicable?

I believe you made a small algebra error in that this:

$c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 2\sqrt{2}$

should be:

$c^2\cos^2{(\phi)} + c^2\sin^2{(\phi)} &= 17-12\sqrt{2} + 8$

thanks

Yes you are correct, that was a typo.

Did you manage to find any values for $\displaystyle \phi$ ?

**VinceW** · August 30th 2011, 11:53 AM

Originally Posted by Prove It

Did you manage to find any values for $\displaystyle \phi$ ?

Yes, that was last step was obvious after all the previous steps:

$\phi = \arccos \frac{3 - 2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} = \arcsin \frac{2\sqrt{2}}{\sqrt{25 - 12\sqrt{2}}} \approx 1.510$

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